Ok. One more attempt! Solution: Let I=∞∫−∞e−x2dx Since x is a dummy variable we can also write: I=∞∫−∞e−y2dy By multiplying the two together, we get: I2=∞∫−∞e−x2dx∞∫−∞e−y2dy=∞∫−∞∞∫−∞e−(x2+y2)dxdy Think of the x and y parameters as Cartesian coordinates, so this double integra;l is the integral over all the plane of e−(x2+y2) Change to polar coordinates : x→rcosθ,y→rsinθ,dxdy→rdrdθ to get: I2=2π∫0dθ∞∫0re−r2dr=2π∞∫0re−r2dr=π∞∫0e−r2dr2=π∞∫0e−udu=π Therefore, since I2=π we have I=√π or ∞∫−∞e−x2dx=√π
Ok. One more attempt!
ReplyDeleteSolution:
Let I=∞∫−∞e−x2dx
Since x is a dummy variable we can also write: I=∞∫−∞e−y2dy
By multiplying the two together, we get:
I2=∞∫−∞e−x2dx∞∫−∞e−y2dy=∞∫−∞∞∫−∞e−(x2+y2)dxdy
Think of the x and y parameters as Cartesian coordinates, so this double integra;l is the integral over all the plane of
e−(x2+y2)
Change to polar coordinates : x→rcosθ,y→rsinθ,dxdy→rdrdθ to get:
I2=2π∫0dθ∞∫0re−r2dr=2π∞∫0re−r2dr=π∞∫0e−r2dr2=π∞∫0e−udu=π
Therefore, since I2=π
we have I=√π
or ∞∫−∞e−x2dx=√π
Test LaTex
ReplyDelete1. I=∞∫−∞e−x2dx
With dollar signs
2. $I = \int\limits_{ - \infty }^\infty {{e^{ - {x^2}}}dx} $
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3. `I=∞∫−∞e−x2dx`
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ReplyDelete4. `$I = \int\limits_{ - \infty }^\infty {{e^{ - {x^2}}}dx} $`
Curly brackets test
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I=∞∫−∞e−x2dx
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